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\title{微分方程数值解\ 第9次作业}
\author{李之琪 22235056}

\begin{document}
\maketitle
\section*{3.8.1}
令$E(t) = \left\|u(\cdot,t) \right\|^2$，我们略去$(\cdot,t)$，则有
\begin{eqnarray}
  \begin{aligned}
    \frac{\text{d} E}{\text{d} t} &= (u_t,u) + (u,u_t)\\
    &=((Bu)_x + Bu_x + Cu ,u) + (u,(Bu)_x + Bu_x + Cu)\\
    &=-(Bu,u_x) + (Bu_x,u) + (Cu,u) - (u_x, Bu) + (u, Bu_x) -
    (Cu,u)\\
    &=-(Bu,u_x) + (Bu_x,u) - (Bu_x, u) + (Bu, u_x) = 0.
  \end{aligned}
\end{eqnarray}
第二步用到了条件$u_t = (Bu)_x + Bu_x + Cu$，第三步用到了分部积分和$C$
的skew-Hermitian性质，第四步用到了$B$的Hermitian性质。因此
\begin{eqnarray}
  \begin{aligned}
    \left\|u(\cdot,t) \right\|^2 = E(t) = E(0) = \left\|u(\cdot,0) \right\|^2.
  \end{aligned}
\end{eqnarray}
\section*{3.8.3}
题中所述的Euler equations如下：
\begin{eqnarray}
  \begin{aligned}
    u_t + \frac{a^2(R)}{R}\rho_x &= \frac{\mu + \mu'}{R}u_{xx},\\
    \rho_t + Ru_x &= 0.
  \end{aligned}
\end{eqnarray}
我们将其重写为
\begin{eqnarray}
  \begin{aligned}
    \begin{bmatrix}
 u\\
 \rho
\end{bmatrix}_t =
\begin{bmatrix}
 \frac{\mu + \mu'}{R} & 0\\
 0 & 0
\end{bmatrix}
\begin{bmatrix}
 u\\
 \rho
\end{bmatrix}_{xx}+
\begin{bmatrix}
 0 & -\frac{a^2(R)}{R}\\
 -R & 0
\end{bmatrix}
\begin{bmatrix}
 u\\
 \rho
\end{bmatrix}_{x}.
  \end{aligned}
\end{eqnarray}
我们知道，对于这样一个对流-扩散系统，系统的相位平移是由对流项决定的，
因此我们只需要考虑对流项的系数矩阵
\begin{eqnarray}
  \begin{aligned}
\begin{bmatrix}
 0 & -\frac{a^2(R)}{R}\\
 -R & 0
\end{bmatrix}
  \end{aligned}
\end{eqnarray}
的行为。注意到该矩阵的特征值为$\lambda = \pm a(R)$，故题中的结论成立。
\end{document}

